Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Show that $\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}=2$
Find $\displaystyle\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}$.
I found that, by calculator, it is actually $\bf{2}$.
Methods to denest something like $\sqrt{a+b\sqrt c}$ seems to be useless here, what should I do?
| We have
$\displaystyle 2 + \frac{10}{9} \sqrt{3} = 1 + \sqrt{3} + 1 + \frac{1}{9} \sqrt{3}$
$\displaystyle = 1 + \frac{3}{\sqrt{3}} + \frac{3}{\sqrt{3}^2} + \frac{1}{\sqrt{3}^3}= \bigl(1 + \frac{1}{\sqrt{3}}\bigr)^3$
Similarily, we have $2-\frac{10}{9} \sqrt{3} = \left(1-\frac{1}{\sqrt{3}}\right)^3$. Hence
$\displaysty... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "23",
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Partial fraction $\frac{x}{(1+x)^2}$ How do you compute $\dfrac{x}{(1+x)^2}$ using partial fractions? The reason I ask is because when I try to solve it I keep getting an impossible $A, B$.
$A(1+x) + B(1+x) = x$
$A + Ax + B + Bx = x$
$(A+B)x = 1$
$(A+B) = 0$
However a practice problem I'm working on implies this can be... | $$\frac{x}{(1 + x)^2} = \frac{A}{1 + x} + \frac{B}{(1 + x)^2}$$
$$x=A(1+x)+B $$
put $x=-1$ gives $B=-1$
and compare coefficient of x from both side gives $A=1$
so answer is $$\frac{x}{(1 + x)^2} = \frac{1}{1 + x} - \frac{1}{(1 + x)^2}$$
| {
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"answer_id": 2
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Proving the inequality $\tan(1)\le\sum_{k=1}^{\infty} \frac{\sin(1/k^2)}{\cos^2 (1/(k+1))}$ How am I supposed to prove this inequality?
$$\tan(1)\le\sum_{k=1}^{\infty} \frac{\sin\left(\frac{1}{k^2}\right)}{\cos^2 \left(\frac{1}{k+1}\right)}$$
Jordan inequality might be an option but led me nowhere. What else to try?
| If $k \in \mathbb{N}$, $k \geq 2$ and $x \in [0, \sqrt{\frac{1}{2}}]$ then
$$\frac{\partial}{\partial x}\sin(\frac{x}{k}) \cos(x) = \frac{1}{k} \cos(\frac{x}{k}) \cos(x) - \sin(\frac{x}{k}) \sin(x) \geq \frac{1}{2 k} - \frac{x^2}{k} \geq 0
$$
so $\sin(\frac{x}{k})\cos(x)$ is non-decreasing on this interval. In particul... | {
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Can an odd perfect number be divisible by $825$? I know that an odd perfect number cannot be divisible by $105$. I wonder if that's also the case for $825$.
| No it cannot.
Let $n=\prod p_i^{\alpha_i}=2^{\alpha_1}3^{\alpha_2}5^{\alpha_3}\cdots$ where $p_i$ is the $i$th prime number and let $S(n)$ be the sum of the divisors of $n$.
Suppose $n$ is an odd perfect number divisible by $825=3\cdot 5^2\cdot 11$, then $S(n)=2n$ and $\alpha_1=0,\alpha_2\ge 1,\alpha_3\ge 2,\alpha_5 \g... | {
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How to prove $(\frac{n+1}{e})^nHow to prove $$\left(\frac{n+1}{e}\right)^n<n!<e\left(\frac{n+1}{e}\right)^{n+1}$$ without integrating method? In fact we could prove this by noticing that
$$i<x<i+1\Rightarrow \ln i<\ln x<\ln (i+1),$$ then integrating it.
| Induct - the base case $n = 1$ is trivial. We will show both sides of the equality by looking at ratios; indeed, $\displaystyle\frac{(n+2)^{n+1} e^n}{e^{n+1} (n+1)^n} = \frac{1}{e}\cdot\left(1+\frac{1}{n+1}\right)^{n} \cdot(n+2) < \frac{1}{1+1/(n+1)} \cdot (n+2) = n+1$ since $\left(1 + \frac{1}{n+1}\right)^{n+1} < e.$ ... | {
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Expand: $e^{\sin z}$ at $z = 0$ I need hints to find the Laurent expansion of $\displaystyle e^{\sin z}$ at $z = 0$ in a simpler way. I am getting double series which I can't simplify.
ADDED:: Can it be simpler that the below? I also need to find the radius of convergence.
$$\Large e^{\sin z} = e^{\sum_{k = 0}^\infty \... | Exponential formula.
Supose
$$ f(x)=a_1 x+{a_2 \over 2}x^2+{a_3 \over 6}x^3+\cdots+{a_n \over n!}x^n+\cdots.$$
Then
$$\exp f(x)=e^{f(x)}=\sum_{n=0}^\infty {b_n \over n!}x^n,$$
where
$$b_n=\sum_{\pi=\left\{\,S_1,\,\dots,\,S_k\,\right\}} a_{\left|S_1\right|}\cdots a_{\left|S_k\right|},$$
and the index $\pi$ runs through... | {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate $\lim_{x\to 1}\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}}$ when $x>1$? Numerically, it looks that the limit is
$$\lim_{x\to 1}\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}} = \frac{1}{\sqrt{2}}\log(1 - x) + cte $$,
but I have not been able to demonstrate it ana... | Here is how you advance, compute the Taylor series of $\frac{1}{\sqrt{x-y}}$ at the point $x=1$
$$ \frac{1}{\sqrt{x-y}} = {\frac {1}{\sqrt {1-y}}}-\frac{1}{2}\, \left( 1-y \right) ^{-3/2} \left( x-1
\right) +O \left( \left( x-1 \right) ^{2} \right) $$
$$ \implies \frac{1}{\sqrt{x-y}} \sim {\frac {1}{\sqrt {1-y}}}. $$... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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For $(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3$, compute $x+y$ . I am trying to find $x+y$ given that
$$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3.$$
It is the radicals in $\sqrt{x^2 +3}, \sqrt{y^2+3}$ that is bugging me. I tried to expand the left hand side
$$ xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)}$$
and see if... | Alt. hint: following up on OP's $^{\tiny\text{(?)}}\,$ approach, with the missing $\,=3\,$ appended...
I tried to expand the left hand side: $$\;xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)} = 3 \tag{1}$$
Before that, multiplying the original equation by $\,\left(x\color{red}{-}\sqrt{x^2+3}\right)\lef... | {
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"url": "https://math.stackexchange.com/questions/391362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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What is $\lim_{n \rightarrow \infty}\sum\limits_{k = 1}^n\frac{k}{n^2}$? We have $$\dfrac{1+2+3+...+ \space n}{n^2}$$
What is the limit of this function as $n \rightarrow \infty$?
My idea:
$$\dfrac{1+2+3+...+ \space n}{n^2} = \dfrac{1}{n^2} + \dfrac{2}{n^2} + ... + \dfrac{n}{n^2} = 0$$
Is this correct?
| Your method is wrong. Consider the first values of the expression:
*
*for $n=1$ you get $1$;
*for $n=2$ you get $1/4+2/4=3/4$;
*for $n=3$ yoy get $1/9+2/9+3/9=6/9=2/3$.
One could make the conjecture that these terms are always bigger than $1/2$, which can be proved by induction. Let
$$
f(n)=\frac{1}{n^2}\sum_{i=... | {
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"timestamp": "2023-03-29T00:00:00",
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Integration of a rational function from +/- infinity I am trying to calculate the integral
$$\int_{-\infty}^{\infty}{\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}}dx$$
where $\{a, b, c\}\in \mathbb{R}$. I have looked in a table of integrals for rational functions, but with no luck. Is there a smart trick I can utilize?... | Hint: Use partial fraction to write the integral as
$$ {\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}} = \frac{A+Bx}{b^2 + (a+x)^2} + \frac{C+D x}{1+c(a-x)^2}, $$
then you can write it as
$$ \frac{A}{b^2 + (a+x)^2} + \frac{C}{1+c(a-x)^2} + \frac{Bx}{b^2 + (a+x)^2} + \frac{D x}{1+c(a-x)^2}, $$
and finally make the chang... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the maximal $t$ satisfying a family of inequalities Given $c \in (0,1)$, find the maximal positive $t$ satisfying the following:
$$\forall n \in \{1,2,\ldots \}: 1+\frac{c}{n+(1-c)} \le \left(1+\frac{1}{n+t}\right)^{c}$$
My progress thus far:
*
*A special case is $c = \frac{1}{2}$: the inequality simplifies ... | The correct $t$ is $\frac{1-c}{2}$, as I was able to guess in the comments using vadim123's calculations.
Proof: define $f(x)=\ln(1+\frac{1}{\frac{1}{x}+t})^c - \ln(1 + \frac{c}{\frac{1}{x}+1-c})$. We are given that $f(\frac{1}{n}) \ge 0$ for any $n \ge 1$.
Simplifying $f$, we find $f(x)=c \ln(1+\frac{x}{1+tx})-\ln(1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/392781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find eigenvectors of $A$ Let $A=\begin{pmatrix} 0&0&0&0\\1&0&0&-2\\0&1&0&1\\0&0&1&2 \end{pmatrix}$
I find eignevalues $-1,0,1,2$ and $0$ has multiplicity of $2$.
But I only find one eigenvector of $0$ and all the other eigenvalues have eigenvectors, so anyway $A$ has four eigenvectors and diagonalizable. But I think $... | $$\begin{align} \det(A-xI) &= \det(\begin{pmatrix}-x&0&0&0\\1&-x&0&-2\\0&1&-x&1\\0&0&1&2-x\end{pmatrix})\\ &=-x(-x(-x(2-x)-1)-2) \\&=-x(2x^2-x^3+x-2)\\&=-x(-x^2(x-2)+x-2) \\&=-x(x-2)(-x^2+1)\end{align}$$
so $x=0,x=2,x=1,x=-1$ are eigenvalues
then we must find X such that :
$(\begin{pmatrix}0&0&0&0\\1&0&0&-2\\0&1&0&1\\0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/393192",
"timestamp": "2023-03-29T00:00:00",
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Definite integration of a trigonometric function
How to integrate $$\int_0^{\pi/2}\!\dfrac{2a \sin^2 x}{a^2
\sin^2 x +b^2 \cos^2 x}\,dx $$
my first step is $$\frac{2}{a} \int_0^{\pi/2}\!\dfrac{a^2 \sin^2 x}{a^2 +(b^2 - a^2) \cos^2 x}\, dx $$
I would kind of want to do some sort of $u=\cos x$ substitution, to get at $... | $$\displaystyle \int_0^{\dfrac{\pi}{2}} \dfrac{2a \sin^2 x}{a^2\sin^2 x +b^2 cos^2 x} dx$$
$$\displaystyle 2a\int_0^{\dfrac{\pi}{2}} \dfrac{\csc^2x }{a^2\csc^2 x +b^2 \cot^2 x \csc^2 x} dx$$
$$\displaystyle 2a\int_0^{\dfrac{\pi}{2}} \dfrac{\csc^2x }{(\cot^2 x+1)(a^2+b^2\cot^2 x)} dx$$
Let $u=\cot x$
$$\displaystyle 2a\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/393266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $x^3-x^{-3}$ given that $x^2+x^{-2} = 83$
If $x>1$ and $x^2+\dfrac {1}{x^2}=83$, find the value of the expression$$x^3-\dfrac {1}{x^3}$$
a) $764$
b) $750$
c) $756$
d) $760$
In this question from given I tried to approximate the value of $x$ which should just above to 9 then I tried to calculate the ... | First, notice that
$$\left(x-\frac{1}{x}\right)^2=x^2-2+\frac{1}{x^2}=83-2=81$$
Then, use the difference of two cubes formula:
$$x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)\left(x^2+1+\frac{1}{x^2}\right)=9\cdot(83+1)=756$$
We take the positive root of $81$ because $x>1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Integral of a rational function: Proof of $\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}= {{\pi}\over{2}}$? I suspect that
$$\sqrt{C}\,\int_{0}^{+\infty }{{{y^2}\over{y^2\,C+y^4-2\,y^2+1}}\;\mathrm dy}=
{{\pi}\over{2}}$$
for $C>0$.
I tried $C=1$, $C=2$, $C=42$, and $C=\frac{1}{1000}$ wit... | So, you can start out by writing the denominator like this: $ y^4+y^2(C-2)+1=(y^2-\frac{2-C-\sqrt{C(C-4)}}{2})(y^2-\frac{2-C+\sqrt{C(C-4)}}{2})$. This is simply the solution of a quadratic equation (ugly, but it will do). So, now, we want to simplify this expression somehow, like by writing:
$ \frac{y^2}{(y^2-\frac{2-C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What would be the value of $a$ and $b$ in following rational expression? If $(5 + 2\sqrt{3})/(7 + \sqrt{3}) = (a - \sqrt{3b})$,
How do I find the value of $a$ and $b$ where $a$ and $b$ are rational numbers?
| $$(5 + 2\sqrt{3}) = (7 + \sqrt{3})(a - \sqrt{3b})$$
$$=7a-7\sqrt b\sqrt 3 +a\sqrt 3-3\sqrt b$$
compare the coefficients of $\sqrt 3$ of both sides
we get$$7a-3\sqrt b=5$$ and $$a-7\sqrt b=2$$
solve for a and b
$a=29/46$ and $\sqrt b=-9/46$
| {
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"timestamp": "2023-03-29T00:00:00",
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integration by substitution, using $\;t = \tan \left(\frac 12 x\right)$ $\displaystyle\int_0^\frac{\pi}{2}\frac{1}{2-\cos x} \, dx$ using the substitution $t=\tan\frac{1}{2}x$
*
*$x=2\tan^{-1}t$
*$\dfrac{dx}{dt}=\dfrac{2}{1+t^2}$
*$dx=\dfrac{2}{1+t^2}\,dt$
*$\displaystyle\int_0^1 \left(\frac{1}{2-\cos x... | If $x=2\arctan t$ then $\cos x = \cos\left(2\arctan t\right)$. Use the fact that $\cos(2u)=\cos^2u-\sin^2u$. So you get
$$
\cos x = \cos^2\left(\arctan t \right) -\sin^2\left(\arctan t \right)
$$
If $\varphi=\arctan t$ then $\dfrac t1=t = \tan\varphi=\dfrac{\text{opposite}}{\text{adjacent}}$, so $\cos\varphi=\dfrac{\... | {
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"timestamp": "2023-03-29T00:00:00",
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Which methods to use to integrate $\int{\frac{x^4 + 1}{x^2 +1}}\, dx$ I have this integral to evaluate:
$$\int{\frac{x^4 + 1}{x^2 +1}}\, dx$$
I have tried substitution, trig identity and integration by parts but I'm just going round in circles.
Can anyone explain the method I need to work this out?
| Let's try some imaginary stuff.
$\dfrac{x^4}{x^2+1}=\dfrac{Ax^3}{x+i}+\dfrac{Bx^3}{x-i} \implies A=B= \dfrac{1}{2}$
$\dfrac{x^4+1}{x^2+1}= \dfrac{1x^3}{2(x+i)}+\dfrac{1x^3}{2(x-i)}+\dfrac{1}{x^2+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/397298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
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Solve equation $\sqrt{s+13} - \sqrt{7-s} = 2$ Solve the equation
$$\sqrt{s+13}-\sqrt{7-s} = 2$$
I moved the $-\sqrt{7-s}$ to the right side
Thus, I had
$$\sqrt{s+ 13} = 2 +\sqrt{7-s}$$
I then squared both sides
$$\sqrt{s+ 13}^2 = \left(2 +\sqrt{7-s}\right)^2$$
Using the product formula $(x + y)^2 = x^2 + 2xy + y^2$
... | You're almost at the end result:
After achieving $ 2s + 2 = 4 \sqrt{7-s} $, square both sides again to generate a quadratic equation.
$$ 4 ( s + 1 )^2 = 16 ( 7 - s ) = 4 (s^2 + 2s+ 1)$$
Solving which is quite easy.
$$\begin{align}
s^2 + 2s + 1 &= 28 - 4s \\
s^2 + 6s - 27 &= 0 \\
(s + 9) (s - 3) &= 0
\end{align}$$
NOTE... | {
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"url": "https://math.stackexchange.com/questions/400153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Help with school solid geometry In the triangular pyramid $MABC$ all side edges equals $1$, $\angle AMB = \angle BMC = 60 ^\circ$, $\angle AMC = 45 ^\circ$.
Find:
1) square of the $\triangle ABC$;
2) dihedral angle on the $AB$ edge;
3) $\rho(M, ABC)$.
Help me, please.
P.S. sorry for my English, I am not a native spea... | Wow the question recalls my old memory about high school.
Actually you know
$$AM = MC = 1$$
and because of the law of cosines
$$cos(\angle AMC) = \frac{\sqrt{2}}{2} = \frac{1^2+1^2-AC^2}{2\cdot 1 \cdot 1}\Rightarrow AC = \sqrt{2-\sqrt{2}}.$$
The area can be computed by Heron's formula.
Notice $\triangle AMC = \trian... | {
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"timestamp": "2023-03-29T00:00:00",
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Need help with integral and change the variable I have
$$\displaystyle \int_0^\pi \sqrt{x} ~\cos x ~dx$$
and I need to make change of the variable $u = \sin x$.
| Hint: Try this if you want:
$$x\sqrt { x } =u\\ x={ u }^{ \frac { 2 }{ 3 } }\\ \sqrt { x } ={ u }^{ \frac { 1 }{ 3 } }\\ dx=\frac { 2 }{ 3 } { u }^{ \frac { -1 }{ 3 } }du\\ \int _{ 0 }^{ \pi } \sqrt { x } \cos x~ dx=\int _{ 0 }^{ \pi }{ u } ^{ \frac { 1 }{ 3 } }\cos { u } ^{ \frac { 2 }{ 3 } }\frac { 2 }{ 3 } {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/404036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$n^5-n$ is divisible by $10$? I was trying to prove this, and I realized that this is essentially a statement that $n^5$ has the same last digit as $n$, and to prove this it is sufficient to calculate $n^5$ for $0-9$ and see that the respective last digits match. Another approach I tried is this: I factored $n^5-n$ to ... | Clearly, $n$ and $n^5$ are of the same parity. Hence, $2 \vert (n^5-n)$.
To check for divisibility by $5$, note that
\begin{align}
n^5 - n & = (n^2+1)n(n+1)(n-1)\\
& = (n^2-4+5)n(n+1)(n-1)\\
& = (n^2-4)n(n+1)(n-1) + 5n(n+1)(n-1)\\
& = (n-2)(n-1)n (n+1)(n+2) + 5n(n+1)(n-1)\\
\end{align}
Clearly, $5 \vert 5n(n+1)(n-1)$. ... | {
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"timestamp": "2023-03-29T00:00:00",
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How do I find the exact value of $\cos\frac{\pi}{12}\cos\frac{5\pi}{12}\cos\frac{7\pi}{12}\cos\frac{11\pi}{12}$? I know that $\cos(6\phi)\equiv32c^6-48c^4+18c^2-1$ where $c=\cos\phi$.
I also know that when $\cos(6\phi)=0$, then $\phi=\frac{k\pi}{12}$ ($k = 1,3,5,7,9,11$).
How do I find the exact value of:
$$\cos\left(\... | From the formulas: $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\\\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$
we have: $\cos(a)\cos(b)=\frac{1}{2}\big(\cos(a+b)+\cos(a-b)\big)$
Apply this formula:
$\cos\left(\frac{\pi}{12}\right)\cos\left(\frac{11\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right)... | {
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"answer_count": 4,
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} |
Limit of a Sequence involving cubic root I succeed in finding the following limit applying binomials and squeeze theorem:
$$\lim(\sqrt{n+1} - \sqrt{n}) = \lim\frac{1}{\sqrt{n+1} + \sqrt{n}} = 0$$ because $0 \leq \frac{1}{\sqrt{n+1} + \sqrt{n}} \leq \frac{1}{\sqrt{n}}$
But I need help because I'm not finding any way to ... | The point is that $\sqrt[3]{n^3-1}$ is awfully close to $n$ when $n$ is large. For a crude estimate, note that
$$\left(n-\frac{1}{n}\right)^3=n^3-3n+\frac{3}{n}-\frac{1}{n^3}\lt n^3 -1$$ if $n\ge 2$. It follows that if $n\ge 2$ then
$$ 0\lt n-\sqrt[3]{n^3-1} \lt n-\left(n-\frac{1}{n}\right)=\frac{1}{n},$$
and the r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/409551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Given $a_1,a_2,...,a_n>0$ where $n\in\mathbb N$$, a_1+a_2+...+a_n=n$. Is this true? $a_1a_2+a_2a_3+...+a_na_1\leq n$
Given $a_1,a_2,...,a_n>0$ where $n\in\mathbb N$$, a_1+a_2+...+a_n=n$. Is this true?
$$a_1a_2+a_2a_3+...+a_na_1\leq n$$
By observing:
When $n=1$, this is trivial;
When $n=2$, $ab\leq(\frac {a+b} 2)^2=1... | The inequality does not hold in general. Indeed, take $a_1=a_2=\frac{n}{2}-1$ and
$a_3=..=a_n=\frac{2}{n-2}.$ Then the left hand side of our inequality is greater that $n^2/4-n+1$ which can be made greater than $n.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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can the following sum be simplified For $n \ge 3$, define
$$f(n) = \sum_{k=3}^n {n \choose k}{k-1 \choose 2}.$$
Is there a closed form expression for $f$?
| We will use the identities
$$
\begin{align}
\sum_{k=m}^n\binom{n}{k}\binom{k}{m}
&=\sum_{k=m}^n\binom{n}{m}\binom{n-m}{k-m}\\
&=\binom{n}{m}2^{n-m}\tag{1}
\end{align}
$$
and
$$
\binom{k-1}{2}=\binom{k}{2}-\binom{k}{1}+\binom{k}{0}\tag{2}
$$
noting that
$$
\binom{k-1}{2}=\left\{\begin{array}{}
0&\text{for }k\in\{1,2\}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/411256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding Maximum Under Constraint Suppose $a$,$b$,$c$ satisfy $a+b+c=1$ and $a$,$b$,$c\in [0,1]$
Find the maximum value of $(a-b)(b-c)(c-a)$
| WOLOG, suppose $c$ is the largest. If $a>b$, then $(a-b)(b-c)(c-a)\le 0$. So we may suppose $a<b<c$. Let $b-a=x\ge 0, c-b=y\ge 0$, then $3a+2x+y=1$, thus $2x+y\le 1$. The problem becomes find maximum value of
$$
xy(x+y), \quad x, y \ge0.
$$
So we may suppose $2x+y=1$, then
$$
xy(x+y)=x(1-2x)(1-x)=:f(x), \quad 0\le x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Find the coefficient using binomial theorem. What is the coefficient of $x^{20}$ in the expression: $$(x+1)^{10}.(x^2 -1)^8$$
| From the binomial theorem you know that
$$(x+1)^{10}=\sum_{k=0}^{10}\binom{10}kx^k$$
and
$$(x^2-1)^8=\sum_{k=0}^8\binom8k(-1)^{8-k}x^{2k}=\sum_{k=0}^8\binom8k(-1)^kx^{2k}\;,$$
so
$$\begin{align*}
(x+1)^{10}(x^2-1)^8&=\left(\sum_{k=0}^{10}\binom{10}kx^k\right)\left(\sum_{k=0}^8\binom8k(-1)^kx^{2k}\right)\\\\
&=\left(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/413342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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If $a$ and $b$ are non-negative real numbers, prove that $ab(a+b) \leq a^2+b^2$.
If $a$ and $b$ are non-negative real numbers, prove that $ab(a+b) \leq a^2+b^2$.
Is is a geometric mean? How to prove it?
| Note that if $a+b\le 2$, then $ab(a+b)\le 2ab$. And we have $2ab\le a^2+b^2$, since $(a-b)^2\ge 0$. So the desired inequality does hold if $a+b\le 2$.
If the condition $a+b\le 2$ is violated, then the inequality can fail. For let $a=b=k$. Then $ab(a+b)=2k^3$ and $a^2+b^2=2k^2$. If $k\gt 1$ then $2k^3\gt 2k^2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove $\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$, given $x+y+z=3$ and $x,y,z\ge0$
Let $x+y+z=3,x,y,z\ge 0$,show that
$$\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$$
Additional information
I have seen the following problem:
$x,y,z>0,x+y+z=3$, prove that
$$\sqrt{x^2+y+2}+\sqrt{y^2+z+2}+\sqrt{z^... | Let's solve
$$ \min \Bigg[\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2} \Bigg]$$
subject to
$$
x+y+z=3 \\
x\ge0 \\
y\ge 0 \\
z\ge 0
$$
Formally, this is done with Kuhn-Tucker condition. Write those down, exploit symmetries, note that non-negativity constraints do not bind.
Solution: $x=y=z=1$. Hence, the minimum valu... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
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Probability of adjacent seating A homework question states:
A room holds two rows of six seats each. Two friends are assigned
randomly to the 12 seats. What is the probability that the 2 friends
sit in adjacent seats?
Note: Friends sitting behind friends don't count. Friends sitting
diagonally adjacent to each o... | The calculation is correct, and efficiently done. There are $\binom{12}{2}$ equally likely ways to select $2$ seats from $12$.
You then counted the "favourables" well, though it was quite unnecessary to list separately the favourables in the first row and the favourables in the second.
Since you did the problem in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/419053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Create Fourier-Series of f(x) = x if 0 < x < Pi and 0 if Pi < x < 2*Pi I tried the following to create the Fourier-series of the function:
$$ f(x) = \begin{cases} x & 0<x<\pi \\ 0 & \pi < x < 2 \pi \end{cases}$$
This is what I tried:
$$a_0 = \frac{1}{\pi}\int\limits_0^\pi x dx = \frac{1}{\pi}\cdot\left(\frac{x^2}{2}... | Plugging in $x=0$ and $x=\pi$ in $\sin (nx)$ indeed gives you $0$.
Plugging in $x=\pi$ in $\cos (nx)$ indeed gives you $(-1)^n$.
But don't forget that plugging in $x=0$ in $\cos (nx)$ gives you $1$.
Also, in the integral for $b_n$, there seems to be a minus sign missing after the integration by parts. Recheck your calc... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solutions for $p$ where $2 p^2 - 1 = q^2$ Consider this equation:
$$2 p^2 - 1 = q^2$$
where $p$ and $q$ are prime.
After vigorous checking, I couldn't find any solutions $p>29.$
Is it so that $p=5,\;p = 29$ are the only solutions?
Edit:
Found solutions $p>29$ with the help this.
The solution for $p,q$ can be expressed ... | There are at least 2 more solutions, you just haven't searched far enough
OEIS A163742 gives solutions $(63018038201, 44560482149)$ and $(19175002942688032928599, 13558774610046711780701)$. It is conjectured that these are the only 4 prime pairs.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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failed application of magicry in Taylor expansion of $1/x^2$ near $x=2$ It's straightforward to find the Taylor expansion for $\frac{1}{x^2}$ near $x=2$ using the the Taylor series definition.
This is turns out to be $\frac{1}{4} - \frac{1}{4} (x-2) + \frac{3}{16}(x-2)^2 + \cdots$
I was trying to be cute by finding the... | Your idea is sound... iff
$$|1-x^2|<1\implies 0<x^2<2\implies |x|<\sqrt2$$
Now, you want to develop around $\,x=2\,$...not good, uh?
My idea:
$$\frac1x=\frac1{2+(x-2)}=\frac12\frac1{1+\left(\frac{x-2}2\right)}=\frac12\left(1-\frac{x-2}2+\frac{(x-2)^2}4-\ldots\right)\implies$$
$$\frac1{x^2}=\frac1x\frac1x=\frac14\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/426583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Product and Sum of Polynomial Roots The ratio of the sum of the roots of the equation, $8x^3+px^2-2x+1=0 $ to the product of the roots of the equation $5x^3+7x^3-3x+q=0 $ is $3:2$. What is the value $\frac{p-q}{p+q}$?
Well I found out the sum of the roots of 1st equation is $-\frac{p}{8}$ and product of roots of secon... | Using Vieta's Formulas,
the sum of the roots of $\ 8x^3+px^2-2x+1=0$ is $\frac{-p}8$
and the product of the roots of $\ 5x^3+7x^3-3x+q=0$ is $\frac {-q}5$
So, $$\frac {\frac{-p}8}{\frac {-q}5}=\frac32\implies \frac pq=\frac{3\cdot 8}{2\cdot 5}=\frac{12}5$$
Now, use Componendo and dividendo
or use $$\frac{p-q}{p+q}=\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given that $xyz=1$ , find $\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}$? I think I solved this problem, but I don't feel $100$ percent sure of my solution. We have:
$xy=\large {\frac 1z}$
$xz=\large \frac 1y$
$yz=\large \frac 1x$
So let's substitute these into our sum:
$\large \frac{1}{1+x+\frac 1z}+\frac{1}{1+... | \begin{align*}
\frac{1}{1+x+xy}+\frac{1}{1+z+xz}+\frac{1}{1+y+yz}&=\frac{1}{1+x+xy}+\frac{xy}{xy(1+z+xz)}+\frac{x}{x(1+y+yz)}\\
&=\frac{1}{1+x+xy}+\frac{xy}{xy+1+x}+\frac{x}{x+xy+1}\\
&=1
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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If $ 5x+12y=60$ , what is the minimum of $\sqrt{x^2+y^2}$? I know this can be easily done by solving for $y$ and substituting, so that you only have to find the minimum value of the parabola $\large x^2 + \left ( \frac {60-5x}{12} \right)^2$ using standard techniques, but is there a less messy way to do this using ineq... | You need to find the foot of the perpendicular to the line $5x + 12y = 60$ that passes through the origin. The line has slope $-\frac{5}{12}$, so any perpendicular has slope $\frac{12}{5}$. The perpendicular passes through the origin, so its equation is $y=\frac{12}{5}x$. Find the intersection of the two lines:
\beg... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove or disprove ${{2a-1\choose a} + {2a-3\choose a-1} + {2a-5\choose a-2} + \dots {1\choose 1}}={2a\choose a+1}$ Prove or disprove $\displaystyle{{2a-1\choose a} + {2a-3\choose a-1} + {2a-5\choose a-2} + \dots {1\choose 1}}={2a\choose a+1}$
This is not homework. I'm trying to prove something related to Catalan number... | This is false. You can see this by testing $a=3$.
$${ 5 \choose 3} + { 3 \choose 2 } + 1 = {6 \choose 4}-1\neq {6 \choose 4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/428012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How many real roots for $ax^2 + 12x + c = 0$? If $a$ and $c$ are integers and $2 < a < 8$ and $-1 < c$, how many equations of the form $$ax^2+12x+c=0$$ have real roots?
| As @DonAntonio give hint a quadratic eqn have real roots iff
$$b^2-4ac\ge0$$
$$144-4ac\ge0\implies ac\le36$$
since $2<a<8$ and $c>-1$ so $a$ will be in set {$3,4,5,6,7$} and $c$ {$0,1,2,3,\cdots$}
from $$ac\le36$$
firstly we will check combination for equality so factors of $36$ in given range
$36=3\times 12,4\times 9,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $abc=1$ and $\prod_{\text{cyc}}(a^2+b^2)=8$, then $(a+b+c)^3\prod_{\text{cyc}}(a^2+b^2)\ge\left (\sum_{\operatorname{cyc}} (a^2b+b^2a) \right) ^3$.
Let $a,b,c>0 $ where $abc=1$ and $(a^2+b^2)(b^2+c^2)(c^2+a^2)=8$. Prove that $$ (a+b+c)^3(a^2+b^2)(b^2+c^2)(c^2+a^2) \ge \big((a^2b+b^2a)(b^2c+c^2b)(c^2a+a^2c)\big) ^3... | By AM-GM, $\left(a^2+b^2\right)\geq 2ab$ and similarly for two other pairs. Hence, $$8=\prod_{\text{cyc}}\,\left(a^2+b^2\right) \geq \prod_{\text{cyc}}\,(2ab)=8\,(abc)^2=8\,.$$
Therefore, the equality holds, implying that $a=b=c=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) $ Could you, please, check if I solved it right.
\begin{align*}
\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
&= \lim_{n \rightarrow \infty} \sqrt{n^2(1 + \frac1n)}
- \sqrt[3]{n^3(1 + \frac1n)})\\
&= \lim_{n \rightarrow ... | You made a mistake when you applied the limit $n \to \infty$ to the $\left(1 + \frac 1n\right)$ terms but not other terms outside. This is not a valid limit application because the whole thing is still in an indeterminate $\infty - \infty$ form.
I can offer you a tedious method that does not require you to estimate the... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 4
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Find the value of $x+y$ if $x^2+y^2+10 = 2\sqrt2x+4\sqrt2y$
If $$x^2+y^2+10 = 2\sqrt{2}x+4\sqrt{2}y$$ then the value of $(x+y)$ is:
a) $4\sqrt{2}$
b) $3\sqrt{2}$
c) $6\sqrt{2}$
d) $9\sqrt{2}$
Please teach me its basics and how to solve it?
| HINT:
We have $$x^2-2\cdot x\cdot \sqrt2 +(\sqrt2)^2+y^2-2\cdot y\cdot 2\sqrt2+(2\sqrt2)^2=10-2-8$$
$$\implies(x-\sqrt2)^2+(y-2\sqrt2)^2=0$$
We know, if $z$ is real , $z^2\ge 0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\frac {a+b+c} 3\geq\sqrt[27]{\frac{a^3+b^3+c^3}3}$.
Given $a,b,c>0$ and $(a+b)(b+c)(c+a)=8$. Show that $\displaystyle \frac {a+b+c} 3\geq\sqrt[27]{\frac{a^3+b^3+c^3}3}$.
Obviously, AM-GM seems to be suitable for LHS.
For RHS, $a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)=(a+b+c)^3-24$, then I don't know what to ... | I found a solution that uses a bit of calculus.
First using the identity you have already found, we figure the inequality is equivalent to
$$\sqrt[3]{\frac{a^3+b^3+c^3+24}{27}} = \frac{a+b+c}{3} \le \sqrt[27]{\frac{a^3+b^3+c^3}{27}}.$$
Setting $x = \frac{a^3+b^3+c^3}{27}$ we find it equivalent to saying
$$(x+\frac89)^9... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How prove this inequality for $a,b,c,d$ are real numbers let $a,b,c,d$ are real numbers,show that
$$2\sqrt{a^2+c^2}+\sqrt{a^2+c^2+3(b^2+d^2)-2\sqrt{3}(ab+cd)}+\sqrt{a^2+c^2+3(b^2+d^2)+2\sqrt{3}(ab+cd)}\ge6\sqrt{|ad-bc|}$$
This problem is creat by China's famous mathematician hua luogeng,http://en.wikipedia.org/wiki/H... | HINT:
$$2\sqrt{a^2+c^2}+\sqrt{a^2+c^2+3(b^2+d^2)-2\sqrt{3}(ab+cd)}+\sqrt{a^2+c^2+3(b^2+d^2)+2\sqrt{3}(ab+cd)}$$ $$=2\sqrt{a^2+c^2}+\sqrt{(a-\sqrt{3}b)^2+(c-\sqrt{3}d)^2}+\sqrt{(a+\sqrt{3}b)^2+(c+\sqrt{3}d)^2}$$ $$\geq 2\sqrt{2|ac|}+\sqrt{2|(a-\sqrt{3}b)(c-\sqrt{3}d)|}+\sqrt{2|(a+\sqrt{3}b)(c+\sqrt{3}d)|}$$
Now,
$$\sqr... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $a^3+b^3+c^3 \geq a^2b+b^2c+c^2a$
Let $a,b,c$ be positive real numbers. Prove that $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$.
My (strange) proof:
$$
\begin{align*}
a^3+b^3+c^3 &\geq a^2b+b^2c+c^2a\\
\sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\
\sum\limits_{a,b,c} a^2 &\geq \sum\limits_{a,b,c} ab\\
a^2+b... | Just assume, wlog $a\leq b\leq c$. Then this equation is all you need:
$$a^3+b^3+c^3=a^2b+b^2c+c^2a+\underset{\geq 0}{\underbrace{(c^2-a^2)(b-a)}}+\underset{\geq 0}{\underbrace{(c^2-b^2)(c-b)}}\geq a^2b+b^2c+c^2a$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Discriminant of a trinomial
Let $b,c \in \mathbb{Z} $ and let $n \in \mathbb{N} $, $n \ge 2. $ Let $f(x) = x^{n} -bx+c$. Prove that $$\hbox{disc} (f(x)) = n^{n }c^{ n-1}-(n-1)^{n-1 }b^{n }.$$
Here $\hbox{disc} (f(x)) = \prod_{i} f'(\alpha_{i} )$ where $\alpha_{1}, \dots, \alpha_{n}$ are the roots of $f(x)$.
After som... | Here is a brute force approach:
$f'(x) = nx^{n-1} - b$, and we want to compute $\prod_i f'(\alpha_i)$. We do this by looking for the minimal polynomial with roots $\alpha_i^{n-1}$.
Note that
$$\begin{array}%
x^n - bx+c = 0 &\Leftrightarrow x(x^{n-1} - b) = -c \\
&\Leftrightarrow x^{n-1} (x^{n-1} - b)^{n-1} = (-1)^{n-1}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maxima/Minima Problem I am working on this Maxima and Minima Problem :
Determine the Max and Min distance of origin from the curve $3x^2+4xy+6y^2=140$
I tried it solving using the lagrange's method of multipliers. I get the following equations
$x+3x\theta+2y\theta=0$
$y+4y\theta+2x\theta=0$
$2z=0$
$3x^2+4xy+6y^2=140$
S... | Those called homogeneous problems can be easily solved
Calling $d^2 = x^2+y^2$ and making $y = \lambda x$ we have
$$
d^2 = x^2(1+\lambda^2)
$$
but from $3x^2+4 x y +6 y^2 = 140$ we obtain
$$
x^2(3+4\lambda+6\lambda^2) = 140
$$
then
$$
d^2 = \frac{140(1+\lambda^2)}{3+4\lambda+6\lambda^2}
$$
now deriving
$$
\frac{d}{d\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/441137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 6
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Prove that $\sum\limits_{k=0}^{n-1}\dfrac{1}{\cos^2\frac{\pi k}{n}}=n^2$ for odd $n$ In old popular science magazine for school students I've seen problem
Prove that $\quad $
$\dfrac{1}{\cos^2 20^\circ} +
\dfrac{1}{\cos^2 40^\circ} +
\dfrac{1}{\cos^2 60^\circ} +
\dfrac{1}{\cos^2 80^\circ} = 40. $
How to prove mo... | In equation $(7)$ of this answer, I compute that
$$
\sum_{k=1}^n\tan^2\left(\frac{k\pi}{2n+1}\right)=n(2n+1)
$$
Thus, with $n=2m+1$,
$$
\begin{align}
\sum_{k=0}^{n-1}\frac1{\cos^2\left(\frac{k\pi}{n}\right)}
&=\sum_{k=0}^{2m}\sec^2\left(\frac{k\pi}{2m+1}\right)\\
&=2\sum_{k=0}^m\tan^2\left(\frac{k\pi}{2m+1}\right)+2(m+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/442715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 5,
"answer_id": 2
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How to show that $f(x,y)$ is continuous. How to show that $f(x,y)$ is continuous.
$$f(x,y)=\frac{4y^3(x^2+y^2)-(x^4+y^4)2x\alpha}{(x^2+y^2)^{\alpha +1}}$$ for $\alpha <3/2$.
Please show me Thanks :)
| The only point where we can have trouble is $(x,y)=0$. To figure out what happens there, use the polar coordinates, as suggested in the comment of davin. We have $x=r\cos\theta$, $y=r\sin\theta$ and therefore
$$x^2+y^2=r^2,\qquad x^4+y^4=r^4\left(\cos^4\theta+\sin^4\theta\right)$$
so that
$$f=r^{3-2\alpha}\left(4\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/444913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Suppose $xyz=8$, try to prove that $\sqrt{\frac{1}{1+x}}+\sqrt{\frac{1}{1+y}}+\sqrt{\frac{1}{1+z}}<2$ Who can help with the following inequality? I can prove it but using some rather ugly approach (e.g. by leveraging the derivative of $\frac{1}{\sqrt{t+1}}+\frac{1}{2}\sqrt{1-\frac{8}{t^2+8}}$ to show this is always les... | It can be proved by contradiction. Set $a=\sqrt{1/(1+x)}$, $b=\sqrt{1/(1+y)}$ and $c=\sqrt{1/(1+z)}$. We have $$8a^2b^2c^2=(1-a^2)(1-b^2)(1-c^2).$$ We need to prove $a+b+c<2$. Assume that $a+b+c\geq 2$, we have
$$1-a^2=(1+a)(1-a)< 2(b+c-1)$$
Since $(1-b)(1-c)>0$, we have $$1-a^2<2bc$$
Then, we have
$$(1-a^2)(1-b^2)(1-c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/446863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 3
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Prove $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 1$ when $(x^2-1)(y^2-1)(z^2-1)=8^3$ Please help to prove this inequality.
Prove $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 1$ when $(x^2-1)(y^2-1)(z^2-1)=8^3$ and each of $x,y,z$ is greater than 1.
Thanks.
| By symmetry, the minimum value of the expression
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
under the constraint $(x^2-1)(y^2-1)(z^2-1)=8^3$ will occur when $x=y=z$. From this, it follows that the minimum occurs when $x=y=z=3$, and the minimum is $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Prove that: $\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}+\frac{1}{d+3}\leq1$ Let $a$, $b$, $c$ and $d$ are non-negative numbers such that $abc+abd+acd+bcd=4.$ Prove that:
$\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}\leq1$
I simplified it and it turns out that it suffices to show
$19\le3(ab+ac+ad+bc+bd+cd)... | Let $a+b+c+d=3u$, $ab+ac+bc+ad+bd+cd=6v^2$, $abc+abd+acd+bcd=4w^3$ and $abcd=t^4$.
Hence, we need to prove that
$$\prod_{cyc}(a+3)\geq\sum_{cyc}(a+3)(b+3)(c+3)$$ or
$$3(ab+ac+bc+ad+bd+cd)+2(abc+abd+acd+bcd)+abcd\geq27$$ or
$$18v^2w^2+8w^4+t^4\geq27w^4$$ or
$$19w^4-18v^2w^2-t^4\leq0$$ or
$$w^2\leq\frac{9v^2+\sqrt{81v^4+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/450940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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show that $\int_{0}^{\infty } \frac{\sin (ax)}{x(x^2+b^2)^2}dx=\frac{\pi}{2b^4}(1-\frac{e^{-ab}(ab+2)}{2})$ show that
$$\int_{0}^{\infty } \frac{\sin (ax)}{x(x^2+b^2)^2}dx=\frac{\pi}{2b^4}\left(1-\frac{e^{-ab}(ab+2)}{2}\right)$$
for $a,b> 0$
I would like someone solve it using contour but also I would to see different... | You can consider employing the residue theorem. The trick is to use a simple contour despite the pole at $z=0$. A way to attack this is to deform the usual semicircular contour at the origin so as not to include the pole.
Consider first the case $a>0$. Rewrite the integral as
$$\frac12 \int_{-\infty}^{\infty} dx \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/451466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How does $\frac{\sqrt{3}}{2} = \sqrt{\frac{3}{4}}$? I'm studying trigonometry and one of the key triangles (60 30 90) has a side of $\frac{\sqrt{3}}{2}$. I understand that it is derived from an equilateral triangle of sides 1, however I can't see how $\frac{\sqrt{3}}{2} = \sqrt{\frac{3}{4}}$ and why $\frac{\sqrt{3}}{2... | $$\sqrt{\frac{3}{4}}=\sqrt{3\cdot4^{-1}}=\sqrt{3\cdot2^{-2}}=\sqrt{3}\cdot2^{-1}=\frac{\sqrt{3}}{2}$$
As to which to use, normally people tend to like to have as little surds as possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/451765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$ Prove that $\tan(6^\circ)\tan(42^\circ) = \tan(12^\circ) \tan(24^\circ)$.
I don't know how to approach this problem. One approach might be to note that $42-6= 24+12$, and then apply the identities for $\tan(A+B)$ and $\tan(A-B)$, but ... | Here is a proof that I got after back calculation.
Let $\theta=36^{\circ}$ Then, one can show that $\theta$ satisfies the following equation $$(\cos \theta +1)(4\cos^2\theta -2\cos\theta-1)=0$$since $\cos 36^{\circ}>0$, $\theta $ satisfies $$4\cos^2\theta-2\cos \theta-1=0\\ \Rightarrow 2\cos 2\theta-2\cos \theta=-1\ \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/455070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\fra... | In the first line you have $4-4.5=\sqrt{(4-4.5)^2}$, which isn't true, because $-0.5\neq 0.5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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"answer_id": 1
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Prove with Induction for $n\in \mathbb{N}$ and $n$ is even for $1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $ I want to prove by indection, for $n\in\mathbb N$ even:
$$1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $$
what I did first is to check the numbers, so if $n$ is even
lets take $n=2$ so $(2\cdot 2-3)^2-(2\cd... | You have proven that the base case works.
Induction Step: Assume that the claim holds true for $n=k$.
It remains to prove the claim true for $n=k+2$. Observe that:
$$ \begin{align*}
&1^2-3^2+5^2-7^2+\dots+(2k-3)^2-(2k-1)^2+(2(k+2)-3)^2-(2(k+2)-1)^2 \\
&= [1^2-3^2+5^2-7^2+\dots+(2k-3)^2-(2k-1)^2]+[(2k+1)^2-(2k+3)^2] \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/458179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to verify these series expansions? We know that $$ x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots (-1)^n \frac{x^{n+1}}{n+1} + \cdots = \log (1+x) $$ whenever $-1<x<1$. What can we say if $x=1$?
That is, does the series $$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdo... | Note that for $x\lt1$
$$
\frac{x^{2k-1}}{2k-1}-\frac{x^{2k}}{2k}\gt0
$$
and
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x^{2k-1}}{2k-1}-\frac{x^{2k}}{2k}\right)
=x^{2k-2}-x^{2k-1}\gt0
$$
Thus, as $x\to1^-$,
$$
f(x)=\sum_{k=1}^\infty\left(\frac{x^{2k-1}}{2k-1}-\frac{x^{2k}}{2k}\right)
$$
is a positive series, increasin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $ p \equiv 1 \pmod{4}$, prove $((\frac{p-1}{2})!)^2 \equiv -1 \pmod {p}$ where p is prime. Characteristics: The fields where $ p \equiv 1 \pmod{4}$ has half the number from 1 to $\frac{p-1}{2}$ both in positive and the negative. There can be paired up such that when multiplied together, they equal $ -(p-1) \equiv +1... | If we start from Wilson's theorem, $(p-1)! \equiv -1 \pmod{p}$ for a prime $p$, then we can write
$$\begin{align}
-1 &\equiv (p-1)! = \prod_{k=1}^{\frac{p-1}{2}} k \cdot \prod_{m = 1}^{\frac{p-1}{2}} (p-m)\\
&\equiv \left(\frac{p-1}{2}\right)! \cdot (-1)^{\frac{p-1}{2}}\prod_{m = 1}^{\frac{p-1}{2}} (m-p)\\
&\equiv (-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/461288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need proofread for deriving quadratic equation formula How to Solve quadratic equation $$ax^{2}+bx+c=0$$ such as $$a \neq 0$$
Divide by a both side from the equation
such as
$$\frac{a}{a}x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$
$$\Rightarrow x^{2}+\frac{b}{a}x + \frac{c}{a} = 0$$
$$\Rightarrow x^{2} + \frac{b}{a}x + (\fr... | You have an error on your second-to-last line.
$$x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$$
is correct, but when you move the $\frac{b}{2a}$ over, you forgot to change its sign. Correctly, this becomes
$$x + \frac{b}{2a} - \frac{b}{2a} = -\frac{b}{2a} \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$$
$$ \implies x = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/461735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find the sum of this : $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+ \sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....$ How to find the sum of the following :
$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+ \sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....+\sqrt{1+\fr... | In addition to AWertheim's answer, in case you haven't studied telescopic series yet, this is how you can calculate $\displaystyle \sum_{n=1}^{1999} \frac{1}{n^{2}+n}$:
First you see that you can write:
$ \displaystyle \frac{1}{n^{2}+n} = \frac{1}{n} - \frac{1}{n+1} $
This technique is called partial fraction decomp... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Given ${x^2-x y+y^2 = 15, x y+x+y = 13}$ find the value of $x^2+6y$ Both x and y are real numbers and x > y .
Given ${x^2-x y+y^2 = 15, x y+x+y = 13}$ find the value of $x^2+6y$ .
I tried solving the second equation to get $y=(13-x)/(x+1)$ and substituted that in equation one to get a quartic equation in x : $x^4+3 x^3... | HINT:
We have , $$x^2-xy+y^2+3(xy+x+y)=15+3\cdot13$$
$$\implies (x+y)^2+3(x+y)-54=0$$
Solve for $x+y$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given a rational number $x$ and $x^2 < 2$, is there a general way to find another rational number $y$ that such that $x^2Suppose I have a rational number $a$ and $a^2 < 2$. Can I find another rational number $B$ such that $a^2<B^2<2$?
Based on the answer to this question, I thought of doing the following:
$$ a^2 < 2 \... | The substitution $$a \leftarrow \frac{a^2+2}{2a} $$ is exactly what you get from Newton's method for the function $f(a)=a^2-2$. This will result in iterations that are greater than $\sqrt{2}$ since $f$ is an increasing convex function on $(0, \infty)$. I will sketch two alternatives, both based on Newton's method, to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/464009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Finding unit's digit in exponentiation Could someone please explain to me how to find the unit's digit in the following expression:
$$7^{95} - 3^{58}$$
| Unit Digit means $\pmod{10}$
Observe that $7\equiv7\pmod{10},7^2=49\equiv9,7^3\equiv9\cdot7\equiv3,7^4\equiv3\cdot7\equiv1$
$3\equiv3\pmod{10},3^2=9\equiv9,3^3=27\equiv7,3^4=81\equiv1$
So, both $3,7$ have a cycle with period $=4$ which can be also confirmed using Euler's Totient Theorem with $\phi(10)=\phi(2)\phi(5)=4$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\lim\limits_{n \to \infty} \frac{1}{n}\sum\limits^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}}$ Find $$\lim_{n \to \infty} \frac{1}{n}\sum^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}}$$
My approach :
$$\lim_{n \to \infty} \frac{1}{n}\sum^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}} =\lim_{n \to \infty} \frac{1}{n^2}\sum^{2n}_{r =1} \fr... | HINT:
Putting $2n=m,$
$$\lim_{n \to \infty} \frac{1}{n}\sum^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}} $$
$$=4\lim_{n \to \infty} \frac{1}{2n}\sum^{2n}_{r =1} \frac{r}{\sqrt{(2n)^2+4r^2}} $$
$$=4\lim_{m\to\infty}\frac1m\sum^m_{r=1}\frac r{\sqrt{m^2+4r^2}}$$
$$=4\lim_{m\to\infty}\frac1m\sum^m_{r=1}\frac {\frac rm}{\sqrt{1+4\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/465075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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what will be the probability? Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?
my approach was,the number of likely combinations are (1,6),(2,6),(3,6),(2,5),(1,5),(3,5),(2... | Alicia is playing the game. She wins if the sum total of values is at least $6$.
If the first toss results in a $4$ or a $5$, she won't get a chance to retoss, and she won't win.
Instant win: If she gets a $6$ on the first toss, then she has won quickly. This has probability $\dfrac{1}{6}$.
We now examine what happens ... | {
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Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$ This is a problem from Spivak's Calculus $3^{rd}$ ed., Chapter I, Problem $6$(d)
Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$.
Proof. Suppose $x^n = y^n$ and $n$ is even. We consider the following cases.
Case 1. $x \geq 0$ and ... | The case $\ x=0,\ y=0\ $ is of the form $\ x=y$.
Now without loss of generality set $\ y=c\cdot x$, which implies $\ c^n=1$, which implies $\ y=\pm\ x$.
| {
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"url": "https://math.stackexchange.com/questions/471946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Determining the general form of $10^x \bmod 210$ While solving a problem I came across solving $10^x\bmod 210$ for various values of $x$. It seems that the values repeat after an interval of 6 for $x\geq4$. Can any one explain how can solve this equation for some value of $x$ and show that it repeats at an interval of ... | $10^x \equiv 0 \mod 2$ and $\mod 5$ for $x \ge 1$.
$10 \equiv 1 \mod 3$ so $10^x \equiv 1 \mod 3$ for all $x \ge 0$.
$10 \equiv 3 \mod 7$ and $3^6 \equiv 1 \mod 7$ so $10^{6k} \equiv 1 \mod 7$ for all $k$. Moreover since $3^2$ and $3^3$ don't work, $10^x \equiv 1 \mod 7$ only for multiples of $6$. If $x, y \ge 1$ and... | {
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The Diophantine equation $x^2 + 2 = y^3$ How to solve the Diophantine equation $x^2 + 2 = y^3$ with $x,y>0$ ?
($x,y$ are integers.)
| Although I prefer an answer without ring theory here is a solution by using the extension $\mathbb{Z}[\sqrt{-2}]$.
All variables in this proof are integers.
$x^2+2=y^3$ factors as $(x+\sqrt{-2})(x-\sqrt{-2})=y^3$.
Since
$1)$ $\mathbb{Z}[\sqrt{-2}]$ is a UFD.
$2)$ the LHS factors into $2$ conjugates which implies that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/473180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the sum of the series $1^2-2^2+3^2-4^2+...-(2n)^2$ Find the sum of the series $$1^2-2^2+3^2-4^2+...-(2n)^2$$
I tried rewriting it as $$\sum_{r=1}^{2n}-1^{n+1}(r^2)$$ but it didn't help.
Also, looked at re-arranging as $$1^2+3^2+5^2+7^2+...+(2n-1)^2$$ and $$-2^2-4-6^2-8^2-...-(2n)^2$$
Still couldn't get to the give... | Note that
$$(2n-1)^2-(2n)^2=1-4n,$$
and therefore your sum is equal to
$$\underbrace{(1+1+\ldots+1)}_{n\;\mathrm{times}}-4\left(1+2+\ldots+n\right)=n-4\cdot\frac{n(n+1)}{2}=-n(2n+1).$$
The only thing one needs to know is the arithmetic progression sum.
| {
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$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):
When $A+B+C = 180^\circ$ \begin{equation*} \tan A + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}
We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\ta... | HINT
$A+B+C = 180$
$A+B = 180 - C$
We'll apply tangent function:
$\tan (A+B) = \tan (180 - C)$
We'll consider the identity:
$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x\tan y}$
$\frac{\tan A + \tan B}{1-\tan A\tan B} = \frac{\tan 180 - \tan C}{1+\tan 180\tan C}$
But $\tan 180 = 0$, therefore, we'll get:
$\frac{\tan A... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 5
} |
How to solve $f_{n}(x)=xf_{n-1}(x)-f_{n-2}(x)$? How to solve the following recurrence relation
$$f_{n}(x)=xf_{n-1}(x)-f_{n-2}(x)$$
with the initial conditions $f_{1}(x)=x, f_{2}(x)=x^2-1$?
The answer is that $f(x+\frac{1}{x})=\frac{1-x^{2n+2}}{x^{n}(1-x^2)}$ (it may be wrong).
| You can use a brute force method. Possibly the least elegant solution, but here goes:
Compute $f_n(x)$ for a few $n$:
$$\begin{aligned}
&f_1(x) = x\\
&f_2(x) = x^2-1\\
&f_3(x) = x^3-2x\\
&f_4(x) = x^4-3x^2+1\\
&f_5(x) = x^5-4x^3+3x\\
&f_6(x) = x^6-5x^4+6x^2-1\\
&f_7(x) = x^7-6x^5+10x^3-4x\\
&f_8(x) = x^8-7x^6+15x^4-10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/481139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Compute the integral $\int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}$
Compute the indefinite integral
$$
\int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}
$$
My Attempt:
$$
\int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}=\int\frac{1}{(x^2-x+1)^{3/2}.\sqrt{\dfrac{x^2+x+1}{x^2-x+1}}}\,dx
$$
Now define $t$ such that $t^2=\dfrac{x^2+x+1}{x^2-x+1}... | According to http://en.wikipedia.org/wiki/Euler_substitution, this integral can have these four approaches to solve:
Approach $1$:
Let $u=x+\sqrt{x^2+x+1}$ ,
Then $x=\dfrac{u^2-1}{2u+1}$
$dx=\dfrac{2u(2u+1)-(u^2-1)2}{(2u+1)^2}du=\dfrac{2u^2+2u+2}{(2u+1)^2}du$
$\therefore\int\dfrac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}$
$=\int\... | {
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"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Real solutions of the equation $x = \sqrt{3-x} \cdot \sqrt{4-x} + \sqrt{4-x} \cdot \sqrt{5-x} + \sqrt{5-x} \cdot \sqrt{3-x}$ Solve the equation
$$x = \sqrt{3-x} \cdot \sqrt{4-x} + \sqrt{4-x} \cdot \sqrt{5-x} + \sqrt{5-x} \cdot \sqrt{3-x},$$
where $x \in \mathbb{R}$.
| I'm assuming we are solving over the reals.
Set $a=\sqrt{3-x}$ and likewise for $b$ and $c$.
Then $x=ab+bc+ca$ and we see $$a^2+x = 3,\qquad b^2+x=4,\qquad c^2+x=5$$
Substituting for $x$, we find $$(a+b)(a+c) = 3,\qquad (b+c)(b+a)=4,\qquad (c+a)(c+b)=5$$
Thus $$(a+b)(a+c)(b+c)=\sqrt{60}$$
This leads to $$a+b=\frac{\sqr... | {
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"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Proving $1+\frac{3}{a+b+c}\geq \frac{6}{ab+bc+ca}$, given $abc=1$ Let a, b, c be positive numbers such that $abc=1$. Prove that $1+\frac{3}{a+b+c}\geq \frac{6}{ab+bc+ca}$
The usual methods do not seem to work, including a substitution $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$ and trying to apply Muirhead's inequali... | $$\Longleftrightarrow (ab+bc+ac)+\dfrac{3(ab+bc+ac)}{a+b+c}\ge 6$$
since by $AM-GM$ inequality
$$ (ab+bc+ac)+\dfrac{3(ab+bc+ac)}{a+b+c}\ge 2\sqrt{\dfrac{3(ab+bc+ac)^2}{a+b+c}}$$
$$\Longleftrightarrow2\sqrt{\dfrac{3(ab+bc+ac)^2}{a+b+c}}\ge 6$$
$$\Longleftrightarrow (ab+bc+ac)^2\ge 3(a+b+c)$$
since by $AM-GM$
$$(ab+bc+ac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/484263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Generating functions and finding coefficient of $x^{3n}$ So I have this homework question I've been trying all day to answer, any help would be appreciated:
Determine using generating functions in how many ways can you choose $3n$ letters out of $\{a, b, c\}$ so that each letter is chosen at most $2n$ times?
| As @user84413's answer tells you, you are after:
$\begin{align*}
[x^{3 n}] (1 + x + x^2 + \dotsb + x^{2 n})^3
&= [x^{3 n}] \left( \frac{1 - x^{2 n + 1}}{1 - x} \right)^3 \\
&= [x^{3 n}] (1 - 3 x^{2 n + 1} + 3 x^{4 n + 2} - x^{6 n + 3}) \cdot (1 - x)^{-3} \\
&= ([x^{3 n}] - 3 [x^{n - 1}]) \sum_{k \ge 0} (-1)^k \bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/485223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How find this $\sum_{n=1}^{\infty}a_{n}$ let $$a_{1}=1,a_{n+1}=\dfrac{1}{a_{1}+a_{2}+\cdots+a_{n}}-\sqrt{2}$$
find the value
$$\sum_{n=1}^{\infty}a_{n}$$
my try:let $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$,then we have
$$S_{n+1}=S_{n}+a_{n+1}=S_{n}+\dfrac{1}{S_{n}}-\sqrt{2}$$
and I find $S_{1}=1,S_{2}=2-\sqrt{2},S_{3}=3-\dfrac... | Very nice problem. Here is my approach:
If the series converges, it converges to $\frac{1}{\sqrt{2}}$ since $a_n\rightarrow 0$. Taken in consideration this, we define $$b_n=a_1+\cdots+a_n-\frac{1}{\sqrt{2}}$$
We want to prove that $b_n$ converges (to zero).
We have
$$
b_{n+1}=b_n+a_{n+1}=b_n-\sqrt{2}+\frac{1}{b_n+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/485816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solutions to $\sqrt{x}+y=6,x^2+y^2=90$ $$\begin{gather}
\sqrt{x}+y=6 \tag{1} \\
x^2 + y^2 = 90 \tag{2}
\end{gather}$$
WE have to solve for $x$ and $y$(Note that 9 is an obvious value of x) My friend asked me this question earlier today, which he said he had made up himself. However, he himself has no idea how to solve... | $$\sqrt x + y = 6\tag{1}$$
$$x^2 + y^2 = 90\tag{2}$$
I'll pursue a "different take" and solve for $y$ in terms of $x$:
First, we can consider the task as one of finding the points of intersection of the curve $\sqrt x + y = 6$, and the circle $x^2 + y^2 = 90$. Plotting both equations in Wolfram Alpha gives us:
It is q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/487360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\lim\limits_{x\to-3}\frac{x^2-9}{2x^2+7x+3}$ I'm going through my textbook, solving problems and am confused because the textbook seems to be giving me the incorrect answer. The question is:
$$ \lim_{x \to -3} \frac{x^2-9}{2x^2+7x+3}$$
The way that I solved this was,
$$\frac{9-9}{...} = \frac{0}{...}$$
Whi... | $\require{cancel}$
Note that both the numerator and denominator evaluate to $0$ at $x = -3$. That means we have an indeterminate limit, which simply means more work needs to be done.
Factor numerator and denominator:
$$\frac{x^2-9}{2x^2+7x+3} = \dfrac{(x + 3)(x - 3)}{(x + 3)(2x + 1)}$$ Cancel the line terms.
$$\frac{(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/490053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Factorize : $x^6 − 10x^3 + 27$ I want to factorize $$x^6 − 10x^3 + 27$$
I tried two methods , first I let $y=x^3 $ and converted it into a quadratic but the solutions are not real .
The second method I tried was getting it to the form $a^3+b^3+c^3-3abc$ but I could only get close ( $(x^2)^3+(x)^3+3^3-3(x^2)(x)(3)$ bu... | In such cases, we can start with
$$(x^2)^3+(ax)^3+b^3-3\cdot x^2\cdot ax\cdot b=x^6+x^3(a^3-3ab)+b^3$$
Comparing with the given expression, $b^3=27, a^3-3ab=-10$
We know, the only real value of $b$ is $3\implies a^3-9a+10=0$
Clearly, $a=2$ satisfies the last equation
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/490643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Question on definite integral How to compute the following:
$$ \int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x(1+x^2)}dx $$
If you could, please, solve it completely, for, I am afraid, a hint won't do for me.
| Let $\displaystyle I(a) = \int_{0}^{\infty}\frac{\tan^{-1}(ax)}{x.(1+x^2)}dx$
Now Diff. both side w.r.to $a$ , We Get
$\displaystyle \frac{dI(a)}{da} = \int_{0}^{\infty}\frac{1}{(1+a^2x^2).x.(1+x^2)}.x.dx$
$\displaystyle \frac{dI(a)}{da} = \int_{0}^{\infty}\frac{1}{(1+x^2).(1+a^2x^2)}dx$
$\displaystyle \frac{dI(a)}{da}... | {
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"url": "https://math.stackexchange.com/questions/491855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving trigonometric inequality with elementary means In a set of olympiad problems, the last,hardest one was:
We know that $\sin(x)<x<\tan(x)$ for every real $x\in (0,\pi/2)$. What is the maximum integer value of $n$ such that holds for every $x\in(0,\pi/2)$
$$\sin(2x)+\tan(2x)>nx$$
By taylor expansion and term b... | I'm assuming that the maximal (integer) $n$ such that
$$\sin (2x) + \tan (2x) > nx$$
holds for $0 < x < \frac{\pi}{4}$ is sought, and not for $\frac{\pi}{2}$. Otherwise no such $n$ exists.
So looking at
$$\frac{\sin (2x)}{2x} + \frac{\tan (2x)}{2x} = \frac{\sin (2x)}{2x}\left(1 + \frac{1}{\cos(2x)}\right) < 1 + \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/492932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to get maximum value of a function with three variables satisfying a condition? The maximum value of the function
$$f(x, y, z) = \left(x -\frac{1}{3}\right)^2 + \left(y - \frac{1}{3}\right)^2 + \left(z - \frac{1}{3}\right)^2$$
subject to the constraints-
$x + y + z = 1; x \ge 0; y \ge 0; z \ge 0$
what is the actual... | This is a smooth function defined in a closed interval and as such would achieve its maximum (and minimum) on the boundary or where the gradient is $0$. Calculus (particularly Lagrangian formulation) is an effective way to handle this. Other ways include using a suitable substitution (if you can find one), using suit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/494713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Limit of $s_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)$ \begin{align*}S_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)\end{align*}
how to calculate the limit $s_n$?
\begin{align*}\lim_{n\to \infty } \, S_n\end{align*}
| Since $$\frac{1}{\sqrt{k}} \ge \frac{2}{\sqrt{k}+\sqrt{k+1}} = 2(\sqrt{k+1}-\sqrt{k}) \ge \frac{1}{\sqrt{k+1}}$$
We find.
$$\begin{align}
& \sum_{k=1}^{n} \frac{1}{\sqrt{k}} \ge 2\sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k}) = 2(\sqrt{n+1}-1) \ge 2\sqrt{n} - 2\\
\text{and}\quad & \sum_{k=1}^{n} \frac{1}{\sqrt{k}} = 1 + \sum_{k=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
How to prove that $\int_0^1 \exp\left(\frac{4x\sqrt{1-x^2}}{\sqrt{8x^2+1}}\right)\sqrt{\frac{1-8x^2+16x^4}{1+7x^2-8x^4}}dx=e-1$ show that
$$\int\limits_0^1 {\exp \left( {\frac{{4x\sqrt {1 - {x^2}} }}{{\sqrt {8{x^2} + 1} }}} \right)} \sqrt {\frac{{1 - 8{x^2} + 16{x^4}}}{{1 + 7{x^2} - 8{x^4}}}} dx = e - 1$$
I think thi... | Let us note that under the change of variables $y=\sqrt{\frac{1-x^2}{1+8x^2}}$ we have $x=\sqrt{\frac{1-y^2}{1+8y^2}}$ and the interval $(0,\frac12)$ is mapped to $(\frac12,1)$ and vice versa. Also,
$$\frac{1-4x^2}{\sqrt{(1-x^2)(1+8x^2)}}dx=\frac{3(4y^2-1)dy}{(1+8y^2)\sqrt{(1-y^2)(1+8y^2)}}$$
Now our integral can be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/497307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
calculate a trigonometric expression related to $\sin(\pi/5)$ I tried to solve a question which led to
$$4 \sin\frac{\pi}{5} \sin\frac{3\pi}{5} = \sqrt{5}$$
But, I got stuck to prove that.
Is there any easy solution?
| Note that
$$\sin{\frac{3 \pi}{5}} = \sin{\frac{2 \pi}{5}}$$
then use a double-angle and triple-angle forumla:
$$\sin{2 x} = 2 \sin{x} \cos{x}$$
$$\sin{3 x} = 3 \sin{x} - 4 \sin^3{x}$$
In this case, $x=\pi/5$. Setting the above two equations equal to each other results in the quadratic equation in question:
$$2 \cos{x}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Product of Permutations Can someone explain how to multiply two permutations? I cannot get myself to understand. Please be very simple and explain each step.
| If you have two permutations displayed in this form:
$$\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 4 & 2 & 5 & 1\end{pmatrix}$$
and
$$\tau = \begin{pmatrix}1 & 2 & 3 & 4 & 5 \\ 5 & 4 & 3 & 2 & 1\end{pmatrix}$$
you can reorder the columns of $\tau$ so that the first line of $\tau$ matches the second line of $\sigm... | {
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"url": "https://math.stackexchange.com/questions/500198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I show that this sequence converges to $0$? For $x \in [0,1]$, let $a_k = k^2x(1-x^2)^k$. How can I show that $\lim_{k \to \infty}a_k = 0$? I don't know what to do because $k^2 \to \infty$ and $(1-x^2)^k \to 0$, so I would have to prove that $\infty \cdot 0 = 0$.
| We see that $\frac{a_{k+1}}{a_k} = \frac{(k+1)^2x(1-x^2)^{k+1}}{k^2x(1-x^2)^{k}} = (1 + \frac{2}{k} + \frac{1}{k^2})(1-x^2)$.
Since $x < 1$, $1 - x^2 < 1$, say equal to $1 - \epsilon$, with $\epsilon > 0$.
Since $\lim_{k \to \infty} (1 + \frac{2}{k} + \frac{1}{k^2}) = 1$, we can take $k$ large enough so that $(1 + \fr... | {
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"source": "stackexchange",
"question_score": "2",
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Integration of $\int_{}^{}\frac{1}{x(1+x)^3}dx $ we got this integration problem
$$\int_{}^{}\frac{1}{x(1+x)^3}dx $$
it seems a fairly simple problem but what i am struggling with it is doing its partital fractions
$\int_{}^{}\frac{1}{x(1+x)^3}dx $ =$\int_{}^{}\frac{a}{x}dx$ +$\int_{}^{}\frac{b}{(1+x)^3}dx$+
$\int_{}... | Ok, your equation is right:
$$a(1 + x)^3 + bx + cx(1 + x) + d(1 + x)^2 = 1$$
Rewriting this:
$$\begin{align}
a(1 + 3x + 3x^2 + x^3) + bx + cx + cx^2 + dx(1 + 2x + x^2) & = 1 \\
a + 3ax + 3ax^2 + ax^3 + bx + cx+cx^2+dx+2dx^2+dx^3& = 1 \\
\end{align}$$
Grouping the terms by the powers of x:
$$\begin{align}
(a+d)x^3 + (3... | {
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"url": "https://math.stackexchange.com/questions/501260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Infinite Series $\sum\limits_{n=1}^{\infty}\frac{1}{4^n\cos^2\frac{x}{2^n}}$ How to prove that
$$\sum_{n=1}^{\infty}\frac{1}{4^n\cos^2\frac{x}{2^n}}=\frac1{\sin^2x}-\frac1{x^2}$$
| Since
$$1 = \sin^2(\theta) + \cos^2(\theta)$$
dividing by $\sin^2(\theta) \cos^2(\theta)$, we get that
$$4\csc^2(2 \theta) = \csc^2(\theta) + \sec^2(\theta)$$
Replace $\theta$ as $\dfrac{x}{2^n}$, we get that
$$\sec^2\left(\dfrac{x}{2^{n+1}}\right) = 4\csc^2\left(\dfrac{x}{2^n}\right) - \csc^2\left(\dfrac{x}{2^{n+1}}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/502676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
how to solve this differential equation $\frac{dy}{dx} = \frac{1+xy}{x(1-xy)}$ by substitution? I've tried with this differential equation $\displaystyle \frac{dy}{dx} = \frac{1+xy}{x(1-xy)}$ , put $u=xy$ then $\displaystyle\frac{du}{dx}=x\frac{dy}{dx}+y$ So,
It will be after editing
$\displaystyle \frac{du}{dx} = \... | Approach $1$:
$\dfrac{dy}{dx}=\dfrac{1+xy}{x(1-xy)}$
$(x-x^2y)\dfrac{dy}{dx}=xy+1$
Let $u=\dfrac{1}{x}-y$ ,
Then $y=\dfrac{1}{x}-u$
$\dfrac{dy}{dx}=-\dfrac{1}{x^2}-\dfrac{du}{dx}$
$\therefore x^2u\left(-\dfrac{1}{x^2}-\dfrac{du}{dx}\right)=x\left(\dfrac{1}{x}-u\right)+1$
$-u-x^2u\dfrac{du}{dx}=2-xu$
$x^2u\dfrac{du}{dx}... | {
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"url": "https://math.stackexchange.com/questions/504057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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When computing $\lim_{x\to -\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}$, I don't get -2. This limit:
$\lim_{x\to -\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}$
is supposed to be equal to -2. My textbook and Wolfram Alpha both state that. However, I can't seem to get same exact result.
Here's what I tried to do:
$$\lim_{x\to -\infty... | We have that : $\frac{2x+7}{\sqrt{x^2+2x-1}}=\frac{x(2+\frac{7}{x})}{\sqrt{x^2(1+\frac{2}{x}-\frac{1}{x^2})}}=\frac{x(2+\frac{7}{x})}{(\left | x \right |)\sqrt{1+\frac{2}{x}-\frac{1}{x^2}}}$. Now use the fact that $\left | x \right |=-x$ for negative $x$ and youre done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Binomial Coefficient Identities I am trying to simplify the following fraction, which I think is equal to 1 but I am not sure.
$$\frac{\frac{\left(\begin{array}{c}
b-1\\
k-1
\end{array}\right)\left(\begin{array}{c}
r\\
n-k
\end{array}\right)}{\left(\begin{array}{c}
r+b-1\\
n-1
\end{array}\right)}}{\frac{\left(\begin{ar... | If we use the identity $\displaystyle\binom{m}{n}=\frac{m}{n}\binom{m-1}{n-1}$, we obtain
$$\displaystyle\frac{\binom{b-1}{k-1}\binom{r+b}{n}}{\binom{b}{k}\binom{r+b-1}{n-1}}=\frac{\binom{b-1}{k-1}\frac{r+b}{n}\binom{r+b-1}{n-1}}{\frac{b}{k}\binom{b-1}{k-1}\binom{r+b-1}{n-1}}=\frac{k}{b}\cdot\frac{r+b}{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Divisors of $9!$ which are perfect square, cube or power $4$ [1] Total no. of positive divisors of $9!$ which are perfect square.
[2] Total no. of positive divisors of $9!$ which are perfect cube.
[3] Total no. of positive divisors of $9!$ which are perfect power 4.
$\underline{\bf{My\; Try}}$:: prime factors of $9! = ... | We use your prime factorization. Note that the perfect squares will have to be of shape $2^{2m} 3^{2n}$, where $m$ and $n$ are non-negative integers, and $2m\le 7$, $2n\le 4$.
For $m$ we can choose any of $0$, $1$, $2$, or $3$ ($4$ choices). For each of these choices, we can let $n=0$, $1$, or $2$. Thus the number of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/505513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove, that $\sin x- a^3\cos x\leq \frac 1 3 \sqrt{1+a^6}$ Let a and $x$ be natural numbers with the property that $\sin x\leq a\cos x$.
Prove that $\sin x- a^3 \cos x\leq \frac 1 3 \sqrt{1+a^6}$.
Again, I'm looking for a second solution. I don't know how to use LaTex and my solution is hard to write here. It would be... | Since $\sin(x)-a\cos(x)\le0$,
$$
\sin(x)-a^3\cos(x)\le(a-a^3)\cos(x)\tag{1}
$$
Since $a\in\mathbb{N}$, $a-a^3\le0$. If $\cos(x)\ge0$, then $(1)$ implies that $\sin(x)-a^3\cos(x)\le0$. So assume wlog that $\cos(x)\le0$. Therefore, $\sin(x)\le a\cos(x)\le0$, and thus,
$$
a\le\tan(x)\tag{2}
$$
Since $x$ is in the third qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/507132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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show that the straight lines $(a^2-3b^2)x^2+8abxy+(b^2-3a^2)y^2=0$ form with the lie $ax+by+c=o$ an equilateral triangle show that the straight lines
$(a^2-3b^2)x^2+8abxy+(b^2-3a^2)y^2=0$ form with the lie ax+by+c=o
an equilateral triangle whose area is $\frac{c^2}{\sqrt{3}(a^2+b^2)}$
is there any other way to solve w... | As I was saying, if we have three constants $F,G,H$ and $$ F x^2 + G x y + H y^2 = 0, $$ we divide through by $x^2$ to arrive at $$ F + G \frac{y}{x} + H \frac{y^2}{x^2} = 0. $$ Or, as I said, using the traditional $m = \frac{y}{x},$ we get
$$ F + G m + H m^2 = 0,$$ or in order
$$ H m^2 + G m + F = 0.$$
We suspect ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/509544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Use induction to prove that $ 1 + \frac {1}{\sqrt{2}} + \frac {1}{\sqrt{3}} .... + \frac {1}{\sqrt{n}} < 2\sqrt{n}$ Use induction to prove that $ 1 + \frac {1}{\sqrt{2}} + \frac {1}{\sqrt{3}} ... + \frac {1}{\sqrt{n}} < 2\sqrt{n} $
My attempt was as follows:
Lets assume the inequality is true for n = k
$S_k = 1 + \fra... | $$2\sqrt{k} + \frac {1}{\sqrt{k+1}} = \frac {2\sqrt{k^2+k}+1}{\sqrt{k+1}} < \frac {2\sqrt{k^2+k+\dfrac14}+1}{\sqrt{k+1}} = \frac{2\left(k+\dfrac12\right)+1}{\sqrt{k + 1}}= 2\sqrt{k + 1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/509840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Infinite Series $\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n}\left\lfloor\frac{\log(n)}{\log(2)}\right\rfloor$ How to prove that
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\left\lfloor\frac{\log(n)}{\log(2)}\right\rfloor=\gamma$$
Can we find a known value for $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\left\lfloor\frac{\log(n)}{\log(... | First show that the sum converges by grouping summands in pairs, so it is enough to check the approximating sums with $n$ running from $1$ to $2^k-1$, say. For these, split the sum into the pieces where $\left\lfloor\frac{\log n}{\log 2}\right\rfloor$ is constant: $$\sum_{n=1}^{2^k-1} \frac{(-1)^n}{n} \left\lfloor \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/509900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Maximize the volume of a rectangular box in the first octant with one vertex in the plane $x+2y+3z=3$ Find the volume of the largest rectangular box in the first octant with the three faces in the coordinate planes and one vertex in the plane $x+2y+3z=3$.
Now I know that $V=xyz$ and I have set $$V=xy\left(1-\frac{x}{3}... | Since $V=xy-\frac{1}{3}x^2y-\frac{2}{3}xy^2$,
$V_{x}=y-\frac{2}{3}xy-\frac{2}{3}y^2=0$ and $V_{y}=x-\frac{1}{3}x^2-\frac{4}{3}xy=0$.
Multiplying by 3 in each equation and dividing by $y$ in the first equation and by $x$ in the second equation (since $x>0$ and $y>0$) gives
$2x+2y=3$ and $x+4y=3$. (Now solve for x and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/510482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to find $\sqrt{1+{4\over x}+{4\over x^2} }$? If $$abx^2 = (a-b)^2(x+1)$$ then what is $$\sqrt{1+{4\over x}+{4\over x^2} }$$
(A) $a+b \over a-b$ (B)$a-b\over a+b$ (C) $a^2+ab$ (d) None
EDIT: What I've done is this:
$$abx^2=(a^2+b^2-2ab)(x+1)$$
=> $$abx^2 - (a^2+b^2+2ab)x - (a-b)^2=0$$
=> $$abx^2 - (a-b)^2x - (... | Hint :
$$\sqrt{1+\frac{4}{x}+\frac{4}{x^2}}=\sqrt{\frac{1}{x^2}(x+2)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/511189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.